Examples

Chapter 7 Class 11 Binomial Theorem
Serial order wise

### Transcript

Question 6 Find the term independent of x in the expansion of (3/2 š„^2 " ā " 1/3š„)^6,x > 0. Calculating general term We know that general term of expansion (a + b)n is Tr + 1 = nCr (a)nār.(b)n For general term of expansion (3/2 š„^2 " ā " 1/3š„)^6 Putting n = 6 , a = 3/2 š„^2 , b = "ā" 1/3š„ Tr + 1 = 6Cr (š/š š^š )^(š ā š) ((āš)/šš)^š = 6Cr (3/2)^(6 ā š) (š„^2 ")" ^((6 ā š)) ((ā1)/3 "Ć" 1/š„)^š = 6Cr (3/2)^(6 ā š) (š„")" ^(2(6 ā š)) ((ā1)/3)^š (1/š„)^š = 6Cr (3/2)^(6 ā š) (š„")" ^(12 ā 2š) ((ā1)/3)^š (š„)^(āš) = 6Cr (3/2)^(6 ā š) ((ā1)/3)^š (š„")" ^(12 ā 2š) (š„)^(āš) = 6Cr (3/2)^(6 ā š) ((ā1)/3)^š (š„")" ^(12 ā 2š ā š) = 6Cr (3/2)^(6 ā š) ((ā1)/3)^š (š„")" ^(12 ā3š) We need to find the term independent of x So, power of x is 0 š„^(12 ā 3š) = x0 Comparing powers 12 ā 3r = 0 12 = 3r 12/3 = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 (3/2)^(6 ā 4) ((ā1)/3)^4 (š„")" ^(12 ā3(4)) T5 = 6C4 (3/2)^2 (1/3^4 ) (š„")" ^(12 ā12) = 6C4 (3^2/2^2 )(1/3^4 ) (š„")" ^0 = 6C4 (1/2^2 )(3^2/3^4 ) (1) = 6C4 (1/2^2 )(1/3^2 ) = 6!/4!(6 ā 4)! (1/4)(1/9) = 6!/4!(2)! (1/4)(1/9) = (6(5)(4)!)/4!(2)! (1/4)(1/9) = (5 )/12 Hence, the term which is independent of x is 5th term = T5 = š/šš